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For a given real symmetric matrix $M$ and nonzero vector $x$, the Rayleigh quotient of $M$ is defined as $R(M,x)=\frac{x'Mx}{x'x}$. I know that for a given matrix, the Rayleigh quotient reaches its minimum value $\lambda_{\min}$ (the smallest eigenvalue of $M$) when $x$ is $v_\min$ (the corresponding eigenvector). Similarly, $R ( M , x )\leq \lambda _{\max }$ and $R(M, v_\max) = \lambda_\max$.

Let $A$ be a non-negative irreducible matrix and $B$ is a matrix obtained from $A$ by adding some ones to it.

how can I conculde that $\lambda^A_{\max}<\lambda_{\max}^B$?

M.Ramana
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1 Answers1

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I suppose that you mean $\rho(A)<\rho(B)$, otherwise your question does not make sense, as the spectra of $A$ and $B$ may contain non-real eigenvalues. If so, the inequality is true and Rayleigh quotient has nothing to do with it. The inequality is a simple consequence of Perron-Frobenius theorem, as discussed here. In general, if $A$ and $B$ are two different nonnegative matrices such that $A+B$ is irreducible, then $\rho(A)<\rho(A+B)$.

user1551
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